A hard disk is connected to a 50 MHz processor through a DMA controller. Assume that the initial set-up of a DMA transfer takes 1000 clock cycles for the processor, and assume that the handling of the interrupt at DMA completion requires 500 clock cycles for the processor. The hard disk has a transfer rate of 2000 Kbytes/sec and average block transferred is 4 K bytes. What fraction of the processor time is consumed by the disk, if the disk is actively transferring 100% of the time?
.
(A) 1.5%
(B) 1%
(C) 2.5%
(D) 10%
Answer: (A)
Explanation: 2000 KB is transferred in 1 second
4 KB transfer is (4/2000)∗1000 ms
= 2 ms
Total cycle required for locking and handling of interrupts after DMA transfer control
=(1000+500) clock cycle =1500 clock cycle
Now, 50 Mhz = 50∗106 = 0.02 microsecond
So, (1500∗0.02)=30 microsecond
30μs for initialization and termination and 2ms for data transfer.
The CPU time is consumed only for initialization and termination.
Fraction of CPU time consumed =30μs/(30μs+2ms)
= 0.015
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